Homework 4: Functions of Several Variables & Probability Solutions

Problem 1: Graph-Function Matching (Optional)

Match the graphs 1-4 to functions i-iv.

functions_to_match

Solution Approach

The matching would involve analyzing:* - Function behavior: Identifying linear, quadratic, exponential, or trigonometric patterns - Domain and range: Understanding where functions are defined and their output values - Critical points: Locating maxima, minima, and inflection points - Symmetry properties: Recognizing even/odd functions or periodic behavior

General Strategy for Graph-Function Matching:

1. Analyze function types:
   • Linear: \(f(x,y) = ax + by + c\) (plane)
   • Quadratic: \(f(x,y) = ax^2 + by^2 + cxy + dx + ey + f\) (paraboloid/saddle)
   • Exponential: \(f(x,y) = e^{g(x,y)}\) (rapid growth/decay)
   • Trigonometric: \(f(x,y) = \sin(g(x,y))\) or \(\cos(g(x,y))\) (periodic)
2. Key visual features:
   • Contour lines: Level curves show function behavior
   • Gradient direction: Steepest ascent direction
   • Critical points: Where all partial derivatives are zero
   • Boundary behavior: Function values at domain edges

Without the specific graphs, detailed matching cannot be provided.

Problem 2: Partial Derivatives

Find the partial derivatives:

1. \(f(x,y)=3x-2y^2\)
2. \(f(x,y)=y^7-2x^3+x^2\)
3. \(f(x,y)=\sin{xy}\)
4. \(f(x,y)=x \cdot \ln y+\dfrac{x}{y}\)

Solutions

1. For \(f(x,y)=3x-2y^2\):
   • \(\frac{\partial f}{\partial x} = 3\)
   • \(\frac{\partial f}{\partial y} = -4y\)
2. For \(f(x,y)=y^7-2x^3+x^2\):
   • \(\frac{\partial f}{\partial x} = -6x^2 + 2x\)
   • \(\frac{\partial f}{\partial y} = 7y^6\)
3. For \(f(x,y)=\sin{xy}\):
   • \(\frac{\partial f}{\partial x} = y\cos(xy)\) (using chain rule)
   • \(\frac{\partial f}{\partial y} = x\cos(xy)\) (using chain rule)
4. For \(f(x,y)=x \cdot \ln y+\dfrac{x}{y}\):
   • \(\frac{\partial f}{\partial x} = \ln y + \frac{1}{y}\)
   • \(\frac{\partial f}{\partial y} = \frac{x}{y} + x \cdot \left(-\frac{1}{y^2}\right) = \frac{x}{y} - \frac{x}{y^2} = \frac{x(y-1)}{y^2}\)

Key Rules Used: - Constant rule: \(\frac{\partial}{\partial x}(c) = 0\) - Power rule: \(\frac{\partial}{\partial x}(x^n) = nx^{n-1}\) - Chain rule: \(\frac{\partial}{\partial x}[g(h(x,y))] = g'(h) \cdot \frac{\partial h}{\partial x}\) - Product rule: \(\frac{\partial}{\partial x}[u \cdot v] = \frac{\partial u}{\partial x} \cdot v + u \cdot \frac{\partial v}{\partial x}\)

Problem 3: Directional Derivatives

Compute the directional derivative at the point \((-1,-1)\) along the vector \(\mathbf{v}=[0.6,\,0.8]\):

1. \(f(x,y) = 3xy\)
2. \(f(x,y) = e^{x-y}\)

Solutions

Formula: The directional derivative of \(f\) at point \((a,b)\) in direction \(\mathbf{u}\) is: \[D_{\mathbf{u}}f(a,b) = \nabla f(a,b) \cdot \mathbf{u}\]

First, verify that \(\mathbf{v}=[0.6,\,0.8]\) is a unit vector: \[\|\mathbf{v}\| = \sqrt{0.6^2 + 0.8^2} = \sqrt{0.36 + 0.64} = 1\] ✓

1. For \(f(x,y) = 3xy\):

   Step 1: Find the gradient

   • \(\frac{\partial f}{\partial x} = 3y\)
   • \(\frac{\partial f}{\partial y} = 3x\)
   • \(\nabla f(x,y) = [3y, 3x]\)

   Step 2: Evaluate at \((-1,-1)\)

   • \(\nabla f(-1,-1) = [3(-1), 3(-1)] = [-3, -3]\)

   Step 3: Compute directional derivative

   • \(D_{\mathbf{v}}f(-1,-1) = [-3, -3] \cdot [0.6, 0.8] = -3(0.6) + (-3)(0.8) = -1.8 - 2.4 = -4.2\)

2. For \(f(x,y) = e^{x-y}\):

   Step 1: Find the gradient

   • \(\frac{\partial f}{\partial x} = e^{x-y}\)
   • \(\frac{\partial f}{\partial y} = -e^{x-y}\)
   • \(\nabla f(x,y) = [e^{x-y}, -e^{x-y}]\)

   Step 2: Evaluate at \((-1,-1)\)

   • \(\nabla f(-1,-1) = [e^{-1-(-1)}, -e^{-1-(-1)}] = [e^0, -e^0] = [1, -1]\)

   Step 3: Compute directional derivative

   • \(D_{\mathbf{v}}f(-1,-1) = [1, -1] \cdot [0.6, 0.8] = 1(0.6) + (-1)(0.8) = 0.6 - 0.8 = -0.2\)

Interpretation: - Negative values indicate the function is decreasing in the given direction - Magnitude shows the rate of change per unit distance

Problem 4: Lake Sevan Navigation

In Lake Sevan, the depth of water at point \((x,y)\) is \(xy^{2}-6x^{2}-3y^{2}\) meters. The ship “Noratus” is at point \((5, 3)\). The first mate suggests sailing north, while the second mate recommends sailing south. Which mate should the captain listen to?

Solution

Note

Plots:

Figure 1: Lake Sevan Depth Visualization

Given: Depth function \(f(x,y) = xy^2 - 6x^2 - 3y^2\) at point \((5,3)\).

Step 1: Find the gradient (direction of steepest increase) - \(\frac{\partial f}{\partial x} = y^2 - 12x\) - \(\frac{\partial f}{\partial y} = 2xy - 6y = 2y(x - 3)\) - \(\nabla f(x,y) = [y^2 - 12x, 2y(x - 3)]\)

Step 2: Evaluate gradient at \((5,3)\) - \(\nabla f(5,3) = [3^2 - 12(5), 2(3)(5 - 3)] = [9 - 60, 6(2)] = [-51, 12]\)

Step 3: Analyze directional derivatives

North direction: \(\mathbf{u}_{north} = [0, 1]\) - \(D_{north}f(5,3) = \nabla f(5,3) \cdot [0,1] = [-51, 12] \cdot [0,1] = 12\)

South direction: \(\mathbf{u}_{south} = [0, -1]\) - \(D_{south}f(5,3) = \nabla f(5,3) \cdot [0,-1] = [-51, 12] \cdot [0,-1] = -12\)

Step 4: Interpretation - North: Directional derivative = +12 (depth increases by 12 m/unit) - South: Directional derivative = -12 (depth decreases by 12 m/unit)

Answer: The captain should listen to the first mate (go North) because sailing north leads to deeper water, which is what the captain wants.

Current depth at \((5,3)\): \(f(5,3) = 5(9) - 6(25) - 3(9) = 45 - 150 - 27 = -132\) meters

Note: The negative depth might indicate this is a relative measurement or the function represents something other than absolute depth.

Problem 5: Local Extrema

Does the following function have local extrema? If so, find them:

1. \(f(x,y) = 3xy\)
2. \(f(x,y) = x^2-xy\)
3. \(f(x, y) = 2x^2 - x^3 - y^2\)

Solutions

Note

Plots:

Figure 2: Functions and Their Critical Points

Method: Find critical points where \(\nabla f = \mathbf{0}\), then use second derivative test.

Second Derivative Test: For critical point \((a,b)\): - \(D = f_{xx}f_{yy} - (f_{xy})^2\) - If \(D > 0\) and \(f_{xx} > 0\): local minimum - If \(D > 0\) and \(f_{xx} < 0\): local maximum
- If \(D < 0\): saddle point - If \(D = 0\): test inconclusive

1. For \(f(x,y) = 3xy\):

Step 1: Find critical points - \(f_x = 3y = 0 \Rightarrow y = 0\) - \(f_y = 3x = 0 \Rightarrow x = 0\) - Critical point: \((0,0)\)

Step 2: Second derivatives - \(f_{xx} = 0\), \(f_{yy} = 0\), \(f_{xy} = 3\) - \(D = (0)(0) - 3^2 = -9 < 0\)

Result: \((0,0)\) is a saddle point (no local extrema)

2. For \(f(x,y) = x^2-xy\):

Step 1: Find critical points - \(f_x = 2x - y = 0 \Rightarrow y = 2x\) - \(f_y = -x = 0 \Rightarrow x = 0\) - Substituting: \(y = 2(0) = 0\) - Critical point: \((0,0)\)

Step 2: Second derivatives - \(f_{xx} = 2\), \(f_{yy} = 0\), \(f_{xy} = -1\) - \(D = (2)(0) - (-1)^2 = -1 < 0\)

Result: \((0,0)\) is a saddle point (no local extrema)

3. For \(f(x, y) = 2x^2 - x^3 - y^2\):

Step 1: Find critical points - \(f_x = 4x - 3x^2 = x(4 - 3x) = 0\) - Solutions: \(x = 0\) or \(x = \frac{4}{3}\) - \(f_y = -2y = 0 \Rightarrow y = 0\) - Critical points: \((0,0)\) and \((\frac{4}{3}, 0)\)

Step 2: Second derivatives - \(f_{xx} = 4 - 6x\), \(f_{yy} = -2\), \(f_{xy} = 0\)

For \((0,0)\): - \(f_{xx}(0,0) = 4\), \(f_{yy}(0,0) = -2\), \(f_{xy}(0,0) = 0\) - \(D = (4)(-2) - 0^2 = -8 < 0\) - Result: Saddle point

Wait, let me recalculate this:

For \((0,0)\): \(D = (4)(-2) - 0^2 = -8 < 0\) → Saddle point

For \((\frac{4}{3}, 0)\): - \(f_{xx}(\frac{4}{3},0) = 4 - 6(\frac{4}{3}) = 4 - 8 = -4\) - \(f_{yy}(\frac{4}{3},0) = -2\), \(f_{xy}(\frac{4}{3},0) = 0\) - \(D = (-4)(-2) - 0^2 = 8 > 0\) - Since \(D > 0\) and \(f_{xx} < 0\): Local maximum

Actually, let me reconsider the analysis for \((0,0)\): - \(D = (4)(-2) - 0^2 = -8 < 0\) → Saddle point ✗

Let me recalculate properly: - At \((0,0)\): \(f_{xx} = 4 > 0\), \(f_{yy} = -2 < 0\), so \(D = 4(-2) - 0 = -8 < 0\) → Saddle point - At \((\frac{4}{3}, 0)\): \(f_{xx} = 4 - 8 = -4 < 0\), \(f_{yy} = -2 < 0\), so \(D = (-4)(-2) - 0 = 8 > 0\)

Since \(D > 0\) and \(f_{xx} < 0\): Local maximum at \((\frac{4}{3}, 0)\)

Final Results: - a) Saddle point at \((0,0)\) - b) Saddle point at \((0,0)\)
- c) Saddle point at \((0,0)\) and local maximum at \((\frac{4}{3}, 0)\)

Problem 6: Optimization - Topless Box (Additional)

You have 12 square meters of cardboard and want to make a topless box. What is the maximum volume your box can have?

Hint: Denote height, length and width by \(x\), \(y\) and \(z\). Express \(z\) by \(x\) and \(y\), then express volume by \(x\) and \(y\).

Solution

Note

Plots:

Figure 3: Box Volume Optimization

Analytical Solution:
For a square base (x = y), optimal dimension is x = y = 2
Optimal height z = (12 - 4)/(4) = 2
Maximum volume = 2 × 2 × 2 = 8 cubic meters

Setup: Let the box have dimensions \(x\) (length), \(y\) (width), and \(z\) (height).

Step 1: Express the constraint Surface area = Bottom + 4 sides = \(xy + 2xz + 2yz = 12\)

Solving for \(z\): \[xy + 2z(x + y) = 12\] \[z = \frac{12 - xy}{2(x + y)}\]

Step 2: Express volume in terms of \(x\) and \(y\) \[V(x,y) = xyz = xy \cdot \frac{12 - xy}{2(x + y)} = \frac{xy(12 - xy)}{2(x + y)}\]

Step 3: Find critical points \[\frac{\partial V}{\partial x} = \frac{y(12 - xy) \cdot 2(x + y) - xy(12 - xy) \cdot 2}{4(x + y)^2} - \frac{xy^2 \cdot 2(x + y)}{4(x + y)^2}\]

This simplifies to: \[\frac{\partial V}{\partial x} = \frac{y(12 - 2xy - y^2)}{2(x + y)^2}\]

Similarly: \[\frac{\partial V}{\partial y} = \frac{x(12 - 2xy - x^2)}{2(x + y)^2}\]

Step 4: Solve \(\nabla V = \mathbf{0}\) From \(\frac{\partial V}{\partial x} = 0\): \(y(12 - 2xy - y^2) = 0\)

Assuming \(y \neq 0\): \(12 - 2xy - y^2 = 0\) … (1)

From \(\frac{\partial V}{\partial y} = 0\): \(x(12 - 2xy - x^2) = 0\)

Assuming \(x \neq 0\): \(12 - 2xy - x^2 = 0\) … (2)

Subtracting (1) from (2): \(x^2 - y^2 = 0\), so \(x = y\) (taking positive values)

Step 5: Find optimal dimensions Substituting \(x = y\) into equation (1): \[12 - 2x^2 - x^2 = 0\] \[12 - 3x^2 = 0\] \[x^2 = 4\] \[x = 2\] (taking positive value)

Therefore: \(x = y = 2\) and \(z = \frac{12 - 4}{4} = 2\)

Step 6: Verify this is a maximum Using the second derivative test or checking boundary behavior confirms this is indeed a maximum.

Answer: The maximum volume is \(V = 2 \times 2 \times 2 = 8\) cubic meters, achieved with a cube of side length 2 meters.

Physical interpretation: The optimal topless box is actually a cube! This makes intuitive sense due to the symmetry of the problem.

Problem 7: Convolution (Additional)

For functions \(f(x)=x^2\) and \(g(x) = \begin{cases} 1 & \text{if } x>0\\ 0 & \text{if } x \le 0 \end{cases}\), find \((f*g)(0)\) where: \[(f*g)(x) = \int_{-1}^1 f(y)g(x-y) \, dy\]

Solution

Given: \(f(x) = x^2\), \(g(x) = \begin{cases} 1 & \text{if } x > 0\\ 0 & \text{if } x \le 0 \end{cases}\), and \(x = 0\).

Step 1: Set up the convolution integral \[(f*g)(0) = \int_{-1}^1 f(y)g(0-y) \, dy = \int_{-1}^1 y^2 g(-y) \, dy\]

Step 2: Analyze \(g(-y)\) \[g(-y) = \begin{cases} 1 & \text{if } -y > 0 \text{ (i.e., } y < 0\text{)}\\ 0 & \text{if } -y \le 0 \text{ (i.e., } y \ge 0\text{)} \end{cases}\]

Step 3: Split the integration interval The interval \([-1,1]\) splits at \(y = 0\): - For \(y \in [-1, 0)\): \(g(-y) = 1\) - For \(y \in [0, 1]\): \(g(-y) = 0\)

Step 4: Evaluate the integral \[(f*g)(0) = \int_{-1}^1 y^2 g(-y) \, dy = \int_{-1}^0 y^2 \cdot 1 \, dy + \int_0^1 y^2 \cdot 0 \, dy\]

\[= \int_{-1}^0 y^2 \, dy + 0 = \left[\frac{y^3}{3}\right]_{-1}^0 = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 - \frac{-1}{3} = \frac{1}{3}\]

Answer: \((f*g)(0) = \frac{1}{3}\)

Interpretation: The convolution measures how much the functions “overlap” when one is shifted. In signal processing and ML, convolution is fundamental for feature detection and filtering operations.

Problem 8: Two Dice Probability

Suppose we roll two fair dice. What is the probability of getting:

1. 2 on each of them
2. at least one 1
3. exactly one 1
   
4. one 1 and one 4
5. 1 on the first die and 4 on the second die

Solutions

Sample space: 36 equally likely outcomes when rolling two dice.

1. 2 on each die:
   • Favorable outcome: \((2,2)\)
   • Probability: \(\frac{1}{36}\)
2. At least one 1:
   • Use complement: \(P(\text{at least one 1}) = 1 - P(\text{no 1s})\)
   • No 1s: Each die shows 2,3,4,5, or 6 (5 choices each)
   • \(P(\text{no 1s}) = \frac{5 \times 5}{36} = \frac{25}{36}\)
   • \(P(\text{at least one 1}) = 1 - \frac{25}{36} = \frac{11}{36}\)
3. Exactly one 1:
   • Case 1: First die is 1, second is not: \((1,2), (1,3), (1,4), (1,5), (1,6)\) → 5 outcomes
   • Case 2: Second die is 1, first is not: \((2,1), (3,1), (4,1), (5,1), (6,1)\) → 5 outcomes
   • Total: 10 outcomes
   • Probability: \(\frac{10}{36} = \frac{5}{18}\)
4. One 1 and one 4:
   • Favorable outcomes: \((1,4)\) and \((4,1)\) → 2 outcomes
   • Probability: \(\frac{2}{36} = \frac{1}{18}\)
5. 1 on first die and 4 on second die:
   • Favorable outcome: \((1,4)\) → 1 outcome
   • Probability: \(\frac{1}{36}\)

Summary: - a) \(\frac{1}{36}\) - b) \(\frac{11}{36}\)
- c) \(\frac{5}{18}\) - d) \(\frac{1}{18}\) - e) \(\frac{1}{36}\)

Problem 9: Colored Pencils

There are 2 red, 5 blue and 6 yellow pencils (total: 13). Two pencils are drawn randomly. Find the probability that both are:

1. red
2. of the same color
3. of different colors
4. not yellow
5. not green

Solutions

Total ways to choose 2 pencils: \(\binom{13}{2} = \frac{13 \times 12}{2} = 78\)

1. Both red:
   • Ways to choose 2 from 2 red: \(\binom{2}{2} = 1\)
   • Probability: \(\frac{1}{78}\)
2. Same color:
   • Both red: \(\binom{2}{2} = 1\)
   • Both blue: \(\binom{5}{2} = 10\)
   • Both yellow: \(\binom{6}{2} = 15\)
   • Total: \(1 + 10 + 15 = 26\)
   • Probability: \(\frac{26}{78} = \frac{1}{3}\)
3. Different colors:
   • Use complement: \(P(\text{different}) = 1 - P(\text{same})\)
   • \(P(\text{different}) = 1 - \frac{1}{3} = \frac{2}{3}\)
4. Not yellow (both red or blue):
   • Non-yellow pencils: 2 red + 5 blue = 7 pencils
   • Ways to choose 2 from 7: \(\binom{7}{2} = 21\)
   • Probability: \(\frac{21}{78} = \frac{7}{26}\)
5. Not green:
   • Since there are no green pencils, all outcomes satisfy this condition
   • Probability: \(\frac{78}{78} = 1\)

Summary: - a) \(\frac{1}{78}\) - b) \(\frac{1}{3}\) - c) \(\frac{2}{3}\) - d) \(\frac{7}{26}\) - e) \(1\)

Problem 10: Dart Throwing

A dart is thrown at a circular target with concentric circles. Circle 1 (innermost) has radius 1m, and each subsequent radius increases by 1m. Find the probability that the dart lands in:

1. circle 1
2. a red circle
3. a yellow circle

dart_throwing

Solutions

Given: Circles with radii 1, 2, 3, 4 meters. Dart lands randomly within the largest circle.

Step 1: Calculate areas - Circle 1 area: \(\pi(1^2) = \pi\) - Circle 2 area: \(\pi(2^2) = 4\pi\)
- Circle 3 area: \(\pi(3^2) = 9\pi\) - Circle 4 area: \(\pi(4^2) = 16\pi\) (total area)

1. Probability of landing in circle 1: \[P(\text{circle 1}) = \frac{\text{Area of circle 1}}{\text{Total area}} = \frac{\pi}{16\pi} = \frac{1}{16}\]

2. Probability of red circle: Assuming circles 2 and 4 are red (pattern: yellow, red, yellow, red)

   Red areas:

   • Circle 2 ring: \(4\pi - \pi = 3\pi\)
   • Circle 4 ring: \(16\pi - 9\pi = 7\pi\)
   • Total red area: \(3\pi + 7\pi = 10\pi\)

   \[P(\text{red}) = \frac{10\pi}{16\pi} = \frac{10}{16} = \frac{5}{8}\]

3. Probability of yellow circle: Assuming circles 1 and 3 are yellow

   Yellow areas:

   • Circle 1: \(\pi\)
   • Circle 3 ring: \(9\pi - 4\pi = 5\pi\)
   • Total yellow area: \(\pi + 5\pi = 6\pi\)

   \[P(\text{yellow}) = \frac{6\pi}{16\pi} = \frac{6}{16} = \frac{3}{8}\]

Verification: \(P(\text{red}) + P(\text{yellow}) = \frac{5}{8} + \frac{3}{8} = 1\) ✓

Summary: - a) \(\frac{1}{16}\) - b) \(\frac{5}{8}\) (assuming pattern) - c) \(\frac{3}{8}\) (assuming pattern)

Note: Parts b) and c) depend on the specific color pattern, which wasn’t provided in the problem.

Problem 11: Coin Tosses - Odd Heads

A fair coin is tossed 5 times. What is the probability of getting an odd number of heads?

Solution

Method 1: Direct Calculation

Odd numbers of heads: 1, 3, 5

\[P(\text{odd heads}) = P(1) + P(3) + P(5)\]

Using binomial probability: \(P(k \text{ heads}) = \binom{5}{k} \left(\frac{1}{2}\right)^5\)

• \(P(1) = \binom{5}{1} \left(\frac{1}{2}\right)^5 = 5 \cdot \frac{1}{32} = \frac{5}{32}\)
• \(P(3) = \binom{5}{3} \left(\frac{1}{2}\right)^5 = 10 \cdot \frac{1}{32} = \frac{10}{32}\)
  
• \(P(5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32}\)

\[P(\text{odd heads}) = \frac{5 + 10 + 1}{32} = \frac{16}{32} = \frac{1}{2}\]

Method 2: Symmetry Argument

For any sequence of fair coin tosses, the probability of getting an odd number of heads equals the probability of getting an even number of heads due to symmetry.

Since these are complementary events: \[P(\text{odd}) + P(\text{even}) = 1\]

By symmetry: \(P(\text{odd}) = P(\text{even})\)

Therefore: \(2 \cdot P(\text{odd}) = 1\), so \(P(\text{odd}) = \frac{1}{2}\)

Answer: \(\frac{1}{2}\)

General Result: For any number of fair coin tosses, the probability of getting an odd number of heads is always \(\frac{1}{2}\).

Problem 12: Conditional Probability - Dice

Two fair dice are rolled. What is the probability of getting 1 on at least one die, given that their sum is even?

Solution

Given information: Sum of two dice is even.

Step 1: Find all outcomes where sum is even Sum is even when both dice show the same parity (both odd or both even).

Both odd: \((1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)\) → 9 outcomes

Both even: \((2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\) → 9 outcomes

Total favorable for condition: 18 outcomes

Step 2: Find outcomes with at least one 1 AND even sum From the “both odd” list above, outcomes with at least one 1: \((1,1), (1,3), (1,5), (3,1), (5,1)\) → 5 outcomes

From the “both even” list, no outcomes contain 1.

Total favorable for both conditions: 5 outcomes

Step 3: Apply conditional probability formula \[P(\text{at least one 1} \mid \text{sum is even}) = \frac{P(\text{at least one 1 AND sum is even})}{P(\text{sum is even})}\]

\[= \frac{\frac{5}{36}}{\frac{18}{36}} = \frac{5}{18}\]

Alternative approach: Among the 18 outcomes with even sum, 5 contain at least one 1. \[P = \frac{5}{18}\]

Answer: \(\frac{5}{18}\)

Problem 13: Playing Cards (Additional)

3 cards are drawn from a deck of 52 cards. What is the probability that the first two cards are queens, and the third one is a diamond ♦?

Solution

Case Analysis: We need to consider whether the queens drawn include the queen of diamonds.

Case 1: Neither of the first two queens is the queen of diamonds

Step 1: Choose 2 queens from the 3 non-diamond queens - Ways: \(\binom{3}{2} = 3\) - Probability: \(\frac{3}{52} \times \frac{2}{51}\)

Step 2: Third card is a diamond (13 diamonds remain) - Probability: \(\frac{13}{50}\)

Case 1 probability: \(\frac{3}{52} \times \frac{2}{51} \times \frac{13}{50}\)

Case 2: One of the first two queens is the queen of diamonds

Step 1: Choose 1 diamond queen and 1 other queen - First card queen of diamonds, second card other queen: \(\frac{1}{52} \times \frac{3}{51}\) - First card other queen, second card queen of diamonds: \(\frac{3}{52} \times \frac{1}{51}\) - Combined: \(\frac{1}{52} \times \frac{3}{51} + \frac{3}{52} \times \frac{1}{51} = \frac{6}{52 \times 51}\)

Step 2: Third card is a diamond (12 diamonds remain, since queen of diamonds already drawn) - Probability: \(\frac{12}{50}\)

Case 2 probability: \(\frac{6}{52 \times 51} \times \frac{12}{50}\)

Step 3: Combine both cases \[P = \frac{3 \times 2 \times 13}{52 \times 51 \times 50} + \frac{6 \times 12}{52 \times 51 \times 50}\]

\[= \frac{78 + 72}{52 \times 51 \times 50} = \frac{150}{132,600} = \frac{1}{884}\]

Answer: \(\frac{1}{884} \approx 0.00113\)

Problem 14: Reading Books (Additional)

There are 15 books: 5 in Armenian, 10 in French. Ruben cannot read French. If he randomly takes 3 books, what is the probability that he can read at least one?

Solution

Complement approach: It’s easier to find the probability that he can read NO books (all French), then subtract from 1.

Step 1: Total ways to choose 3 books from 15 \[\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455\]

Step 2: Ways to choose 3 French books from 10 \[\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\]

Step 3: Probability of choosing all French books \[P(\text{all French}) = \frac{120}{455} = \frac{24}{91}\]

Step 4: Probability of reading at least one book \[P(\text{at least one Armenian}) = 1 - P(\text{all French}) = 1 - \frac{24}{91} = \frac{67}{91}\]

Verification using direct method: - Exactly 1 Armenian: \(\binom{5}{1} \times \binom{10}{2} = 5 \times 45 = 225\) - Exactly 2 Armenian: \(\binom{5}{2} \times \binom{10}{1} = 10 \times 10 = 100\)
- Exactly 3 Armenian: \(\binom{5}{3} \times \binom{10}{0} = 10 \times 1 = 10\) - Total: \(225 + 100 + 10 = 335\) - Probability: \(\frac{335}{455} = \frac{67}{91}\) ✓

Answer: \(\frac{67}{91} \approx 0.736\) or about 73.6%

Bonus Problem 1: Shakespeare’s Monkeys - Infinite Monkey Theorem

A monkey sits at a typewriter and randomly presses keys. The typewriter has 26 letters (a-z), space, and period (28 total keys). What is the probability that the monkey will eventually type all the works of Shakespeare (given infinite time)?

Solution

📹 Video Solution in Armenian

Bonus Problem 2: Bertrand’s Paradox - Random Chords in Circles

Consider two concentric circles with radii 1 and 2. A chord is drawn randomly in the larger circle. What is the probability that this chord intersects the smaller circle?

Solution

📹 Video Solution in Armenian

Bonus Problem 3: Three-Person Duel - Optimal Strategy

Three people (A, B, C) participate in a duel with the following shooting accuracies: - A hits with probability 0.3 - B hits with probability 1 (never misses) - C hits with probability 0.5

They shoot in order A → B → C → A → … until only one survives. What is A’s optimal strategy?

Solution

📹 Video Solution in Armenian